Answer:
Percent yield = 57.8 %
Theoretical yield = 11.781 g
Explanation:
Given data:
Mass of CaO produced = 6.81 g
Mass of CaCO₃ react = 20.7 g
Theoretical yield = ?
Percent yield = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃ :
Number of moles = mass/molar mass
Number of moles = 20.7 g/ 100.1 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of CaCO₃ with CaO.
CaCO₃ : CaO
1 : 1
0.21 : 0.21
Theoretical yield of CaO:
Mass = number of moles × molar mass
Mass = 0.21 mol × 56.1 g/mol
Mass = 11.781 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) × 100
Percent yield = (6.81 g/ 11.781 g) × 100
Percent yield = 57.8 %
Answer:
130 moles of water might be produced by the reaction.
Explanation:
We state the equation for production of water:
2H₂(g) + O₂ (g) → 2H₂O (l)
Moles of hydrogen: 130
Moles of oxygen: 60
We state that the limting reactant is the oxygen
For 2 moles of hydrogen, we need 1 mol of oxygen
For 130 moles of hydrogen, we would need 65 moles, and we only have 60. Ratio is 2:1
1 mol of O₂ can produce 2 moles of water
Then, 60 moles of O₂ will produce (60 . 2) /1 = 130 moles
<span>Answer:
Pb(NO3)2+2HCl=PbCl2+2HNO3
Step 2
Find # of moles of Pb
13.87g Pb(NO3)2 x 1 mol/331.22=.041875 mol Pb(NO3)2
Step 3
Use balanced equation to find weight of Pb(NO3)2
.041875mol Pb x 1 mol Cl2/1mol Pb x 70.90g/1 mol Cl2=2.9689g Cl2
.041875 mol Pb x 207.2/1mol Pb=8.6765g Pb
=11.65g</span>
Answer:
The reaction will absorb energy
Explanation:
A potential energy diagram shows the energy changes that occur during a chemical reaction. It is also called a reaction progress curve. The diagram shows the changes that occur as reactants are converted into products.
The best conclusion according to the energy diagram of the chemical reaction is that the reaction will absorb energy.
Answer:
The equilibrium concentration of hydrogen gas is 0.0010 M.
Explanation:
The equilibrium constant of the reaction = 
}
Moles of hydrogen sulfide = 0.31 mol
Volume of the container = 4.1 L
![[concentration]=\frac{moles}{volume (L)}](https://tex.z-dn.net/?f=%5Bconcentration%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%20%28L%29%7D)
![[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B0.31%20mol%7D%7B4.1%20L%7D%3D0.076%20M)
Initially
0.076 M
At equilibrium
(0.076-2x) 2x x
The expression of an equilibrium constant :
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Solving for x:
x = 0.00051
The equilibrium concentration of hydrogen gas:
![[H_2]=2x=2\times 0.00051 M=0.0010 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D2x%3D2%5Ctimes%200.00051%20M%3D0.0010%20M)
